Question

Стрелок три раза стреляет по мишеням. Вероятность попадания в мишень при одном выстреле равна 0,6. Найдите вероятность того. что стрелок первый раз попал в мишени, а последние два раза промахнулся. Ответ напишите подробно. Решите пожалуйста, очень нужно.

Answer 1

Поскольку стрелок попадает в мишени с вероятностью p=0,6, то промахивается с вероятностью q=1-p=1-0,6=0,4.
События A- "попадание" и Б- "промах" при каждом выстреле независимы.
Вероятность произведения независимых событий равна произведению их вероятностей т.е.
Вероятность=А*Б*Б=0.6*0.4*0.4=0.096
Ответ: 0,096 или 9,6%

Answer 2

Problem Analysis

We are given that a shooter takes three shots at a target, and the probability of hitting the target with a single shot is 0.6. We need to find the probability that the shooter hits the target on the first shot and misses on the next two shots.

Solution

To find the probability of hitting the target on the first shot and missing on the next two shots, we can multiply the probabilities of each event occurring.

Let's denote the probability of hitting the target as p and the probability of missing the target as q. In this case, p = 0.6 and q = 1 - p = 0.4.

The probability of hitting the target on the first shot is p. The probability of missing the target on the second shot is q. The probability of missing the target on the third shot is also q.

Therefore, the probability we are looking for is p * q * q.

Substituting the values, we have: p * q * q = 0.6 * 0.4 * 0.4 = 0.096.

So, the probability that the shooter hits the target on the first shot and misses on the next two shots is 0.096.

Answer

The probability that the shooter hits the target on the first shot and misses on the next two shots is 0.096.